Question

For the reaction, $X_2{Y}_4(l)⟶2X{Y}_2(g)$ at $300K$, the values of $\Delta U$ and $\Delta S$ are $2kcal$ and $20cal{K}^{-1}$ respectively. The value of $\Delta G$ for the reaction is

• A. - 3400 cal
• B. 3400 cal
• C. - 2800 cal
• D. 2000 cal

Answer 1

Answer: (C)

For the reaction $X_2{Y}_4(l)⟶2X{Y}_2(g)$
$\Delta n_g=$ number of gaseous products
- number of gaseous reactants
$\Delta n_g=2-0=2$
Given that, $\Delta U=2kcal$
$\Delta S=20cal{K}^{-1}$
From the formula,
$\Delta H=\Delta U+\Delta n_{g}RT$
Putting the values, we get
$\Delta H=2+\left(\frac{2\times 2\times 300}{1000}\right)$
$=3.2kcal=3.2\times 10^{3}cal$
Also, $\Delta G=\Delta H-T\Delta S$
$=3.2\times 10^3-300\times 20$
$=3.2\times 10^3-6\times 10^3$
$=-2800cal$